Problem: Suppose we have a vector field $f(x, y) = \left( 1, 2 \right)$ and a curve $C$ that is parameterized by $\alpha(t) = (3\cos(t), 3\sin(t))$ for $0 < t < \pi$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Answer: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = \left( 1, 2 \right)$ and $\alpha(t) = (3\cos(t), 3\sin(t))$. $\begin{aligned} &f(\alpha(t)) = (1, 2) \\ \\ &\alpha'(t) = (-3\sin(t), 3\cos(t)) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_0^\pi (1, 2) \cdot (-3\sin(t), 3\cos(t)) \, dt$ Let's solve the integral. $\begin{aligned} &\int_0^\pi (1, 2) \cdot (-3\sin(t), 3\cos(t)) \, dt \\ \\ &= \int_0^\pi -3\sin(t) + 6\cos(t) \, dt \\ \\ &= \left[ 3\cos(t) + 6\sin(t) \right]_0^\pi \\ \\ &= (3 (-1) + 6 (0)) - (3 (1) + 6 (0)) \\ \\ &= -6 \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = -6$.